**Sign of the Quadratic Expression**

**Sign of the Expression ax² + bx + c:**

1. Sign of the expression ax² + bx + c is same as that of ‘a’ for all value of x if b² – 4ac ≤ 0 i, e. if the roots of ax² + bx+ c = 0 are imaginary or equal.

2. If the roots of the ax² + bx + c is real and different i, e. b² – 4ac > 0, the sign of the expression is same as that of ‘a’ if x does not lie between the two roots of the equation and opposite to that of ‘a’ if x lies between the roots of the equation.

3. The expression ax² + bx + c is positive for all real value of x if b² – 4ac < 0 and a > 0.

4. The expression ax² + bx + c has a maximum value when ‘a’ is negative and x = – b/2a. maximum value of the expression = (- b² + 4ac)/4a.

5. The expression ax² + bx + c, has a minimum when ‘a’ is positive and x = -b/2a. minimum value of the expression = (- b² + 4ac)/4a.

**Example 1:** Solve for x, x² – 3x + 2 > 0.

**Solution: **Given x² – 3x + 2 > 0,

Here, sign of co -efficient of x² and expression x² – 3x + 2 > 0 are same due to this x cannot lie between critical point.

For critical point x² – 3x + 2 > 0.

x² – 2x – x + 2 > 0

x (x – 2) -1 (x – 2) > 0

(x -1) (x – 2) > 0

x = 1, 2

Hence, solution of given inequation, x ϵ (- ∞, 1) ∪ (2, ∞).

**Example 2: **Solve for x, x² – 3x + 2 < 0

**Solution: **Given x² – 3x + 2 < 0

Here, sign of co-efficient of x² and expression x² – 3x + 2 are opposite. Due to this x lies between critical points.

For critical points, x² – 3x + 2 < 0,

x² – 2x – x + 2 < 0

x (x – 2) – 1 (x – 2) < 0

(x – 1) (x – 2) < 0

x = 1, 2

Hence, solution of given inequation, x ϵ (1, 2).

**Example 3: **Solve x, – x² + 3x -2 < 0

**Solution:** Given – x² + 3x – 2 < 0

Here, sign of co-efficient of x² and expression – x² + 3x – 2 are same. Due to this x cannot lie between critical points.

For critical points, – x² + 3x – 2 < 0,

– x² + 3x – 2 < 0

x = 1, 2

Hence solution of given inequation, x ϵ (-∞, 1) ∪ (2, ∞).